3 GC + 1 OC question

[Josh779]

Advertisement - Members don't see this ad

Thought I would put these all in one post instead of flooding the forums. I got these wrong in a practice test and am super confused.
The volume of an ideal gas is zero at

A. -273K
B.-273C
C.0C
D.0F
E.212F

Ans:B

and

Which of the following is produced by the electrolysis of 18g of H20?

A. 1 mole O2
B. 2 mole O2
C. 1 mol of H2
D. 2 mole of H2
E. 1/2 mole of H2

Ans: C

and

Consider the equilibrium reaction
2NO(g) + O2(g) <> 2NO2(g); K=2.0

Which of the following is the correct value of K for the reaction
NO2(g)<>NO(g) + 1/2 O2(g)?

A. (1/2.0)^1/2
B. (-2.0)^1/2
C. (2.0/2)
D. (1/2.0)^2
E. (-2.0)^2

Ans:A

How mnay isomeric mono-substitution products are possible for the bromination of the aromatic hydrocarbon naphthalene?

A. 1
B. 2
C. 3
D. 4
E. 5

Thanks! These are the last few things i'm going to cover for the chemistry sections before the big day.

 

[LetsGo2DSchool]

1. Zero volume of a gas is defined at 0 K which is equal to -273 C (Answer B). No such temperature below 0 K so A is incorrect.


2. Write the balanced equation first:

2H2O ---> 2H2 + O2

Since it's talking about electrolysis of 1 mole of H2O (1 mole = 18 g) you can view the equation as:

H2O ---> H2 + 1/2 O2

Thus answer is C.

 

[AlbinoPolarBear]

The volume of an ideal gas is zero at

A. -273K
B.-273C
C.0C
D.0F
E.212F

Ans:B

Given:
Volume = 0 L

We know by Charles' Law that the volume of gas at absolute zero (0K or -273 C) is zero.

 

[herkulease]

1. -273C = 0 kelvin which is absolute zero. the kelvin scale doesn't go into negative.

2.
balanced equation
2H2O -> 2H2 + O2

If you have 18grams of H2O then you have 1 mol of H2O

So

1 mol of H2O X 2 mols H2/2 mol H2O = 1 mol of H2

or

1 mol of H2O X 1 mols O2/2 mol H2O = 1/2 mol O2

the answer only allows for the 1st choice.

3. this questions requires you know 2 things.

the provided K of 2 is the reaction going forward.

the one they want you to determine is in the reverse. so K would be the inverse of the forward reaction. So that's where (1/2) comes from.

Also you will notice that it is also half the amounts so you have to square root it also.

thus you have answer (1/2)^1/2

4. I'm guessing this comes from the paid ada sample. I read about this problem, they provided the wrong structure. the structure provided anthracene 3 rings will get you 3 products. naphthalene which is only 2 rings will get you 2 products.

but don't quote me on that, I don't have the sample so I can't tell you that is indeed true the error that is.

 

[AlbinoPolarBear]

Consider the equilibrium reaction
2NO(g) + O2(g) <> 2NO2(g); K=2.0

Which of the following is the correct value of K for the reaction
NO2(g)<>NO(g) + 1/2 O2(g)?

A. (1/2.0)^1/2
B. (-2.0)^1/2
C. (2.0/2)
D. (1/2.0)^2
E. (-2.0)^2

Ans:A

Just take this question 1 step at a time. There are 2 transformations to get to the final answer.

(1) 2NO + O2 <-> 2NO2, K=2.0
(2) 2NO2 <-> 2NO + O2

K = [P]/[R]

So if we flip P and R, we are gonna get [R]/[P] = 1/K, therefore
(2 )2NO2 <-> 2NO + O2, K=1/2.0

Next, we half the reaction, and this what you have to memorize.

A <->B, K= K
2A <-> 2B, K= K^2
0.5A <-> 0.5B, K = K^0.5

So the final transformation, we half it from 2 to 3.

(2) 2NO2 <-> 2NO + O2, K=1/2.0
(3) NO2 <-> NO + 1/2 O2, K=(1/2.0)^0.5

 

[Josh779]

1. Zero volume of a gas is defined at 0 K which is equal to -273 C (Answer B). No such temperature below 0 K so A is incorrect.


2. Write the balanced equation first:

2H2O ---> 2H2 + O2

Since it's talking about electrolysis of 1 mole of H2O (1 mole = 18 g) you can view the equation as:

H2O ---> H2 + 1/2 O2

Thus answer is C.

Ok so I get the second one now, but the first one i'm still confused on. Isn't it an ideal gas assumption that gas molecules contain no volume under ideal conditions? The whole high temperature low pressure deal...that is a really low temperature to behave ideal...or does the low pressure only apply to the gases not containing any volume...

 

[AlbinoPolarBear]

How mnay isomeric mono-substitution products are possible for the bromination of the aromatic hydrocarbon naphthalene?

A. 1
B. 2
C. 3
D. 4
E. 5

That's naphthalene. There are only 2 ways you can mono-brominate that, without getting identical structures.

 

[AlbinoPolarBear]

Ok so I get the second one now, but the first one i'm still confused on. Isn't it an ideal gas assumption that gas molecules contain no volume under ideal conditions? The whole high temperature low pressure deal...that is a really low temperature to behave ideal...or does the low pressure only apply to the gases not containing any volume...

Ideal gas just means it follows this equation:

PV = nRT

We can rearrange it to have T (which is in K) one side.

T = PV/nR

If V = 0 L, then T = 0 K.

Ideal gas assumes that the molecules contain no volume. When you have high temperature or high pressure, the molecules of gas are more likely to collide and their volumes are more prominent, which will give you inaccurate calculations if you use the PV=nRT equation. There are adjustments to the equation that deals with non-ideal gases that account for their volume and collisions.

 

[LetsGo2DSchool]

Ideal gas just means it follows this equation:

PV = nRT

We can rearrange it to have T (which is in K) one side.

T = PV/nR

If V = 0 L, then T = 0 K.

Da Bear always has my back 👍

 

[Josh779]

Ideal gas just means it follows this equation:

PV = nRT

We can rearrange it to have T (which is in K) one side.

T = PV/nR

If V = 0 L, then T = 0 K.

Ideal gas assumes that the molecules contain no volume. When you have high temperature or high pressure, the molecules of gas are more likely to collide and their volumes are more prominent, which will give you inaccurate calculations if you use the PV=nRT equation. There are adjustments to the equation that deals with non-ideal gases that account for their volume.

oohh jeez, I can't believe I didn't think of that. Thanks man.

 

[Josh779]

How mnay isomeric mono-substitution products are possible for the bromination of the aromatic hydrocarbon naphthalene?

A. 1
B. 2
C. 3
D. 4
E. 5

That's naphthalene. There are only 2 ways you can mono-brominate that, without getting identical structures.

Does the bromine add to the tertiary carbon? or just the two secondary positions..

 

[mh0000]

Does the bromine add to the tertiary carbon? or just the two secondary positions..

No it does not because it cannot eliminate a H to restore aromaticity. C1 is the preferred site of bromination. C2 is plausible, but only as a minor product due to some resonance structures not being as stable as the resonance structures when C1 is brominated.

 

[AlbinoPolarBear]

Does the bromine add to the tertiary carbon? or just the two secondary positions..

You mean the carbons in the middle? It doesn't add there as it will lose aromaticity.

 

[Josh779]

Does the bromine add to the tertiary carbon? or just the two secondary positions..

Is there somewhere that covers this? i.e. Chad or any other source?
Thanks for your help guys! These are starting to make sense.

 

[AlbinoPolarBear]

Is there somewhere that covers this? i.e. Chad or any other source?
Thanks for your help guys! These are starting to make sense.

Hmm.. unfortunately, this is one of the application questions where it's not covered in any test prep material.

 

[rah08e]

Regarding this statement from AlbinoPolarBear...

Next, we half the reaction, and this what you have to memorize.

A <->B, K= K
2A <-> 2B, K= K^2
0.5A <-> 0.5B, K = K^0.5


Why do we have to "Half" the reaction? Is it because the reaction we are finding "K" for is half of the original reaction (K=2)? And since we do, to express the reaction being halved, wouldn't you write "A <-> 0.5B, K= K^0.5?

-Thanks

 

[mh0000]

Regarding this statement from AlbinoPolarBear...

Next, we half the reaction, and this what you have to memorize.

A <->B, K= K
2A <-> 2B, K= K^2
0.5A <-> 0.5B, K = K^0.5


Why do we have to "Half" the reaction? Is it because the reaction we are finding "K" for is half of the original reaction (K=2)? And since we do, to express the reaction being halved, wouldn't you write "A <-> 0.5B, K= K^0.5?

-Thanks

The K is for a reaction with the ratios 2:1:2. We want to find the K for a related( but different reaction) that has 1: 0.5 :1, which is half of the original coefficients. So yes, this does mean taking the sqrt of the original K.

 

[rah08e]

The K is for a reaction with the ratios 2:1:2. We want to find the K for a related( but different reaction) that has 1: 0.5 :1, which is half of the original coefficients. So yes, this does mean taking the sqrt of the original K.

Okay so if the similar but different rxn had a ratio of 4:2:4 instead of 1: 0.5 :1, we would square the K value? What if its 8:4:8?

 

[mh0000]

Okay so if the similar but different rxn had a ratio of 4:2:4 instead of 1: 0.5 :1, we would square the K value? What if its 8:4:8?

First, I should perhaps clarify what I said earlier. I should have said that when you manipulate the coefficients by multiplying it by a certain factor, you are not really "changing" the equilibrium constant (only temperature changes actually change the K value). Each K value describes a very specific reaction with very specific coefficients. So when we change these coefficients, we are just calculating the K value for that describes the new coefficients. So while I wouldn't necessary call it a different reaction, we are dealing with a different K value.

That probably didn;t make too much sense, lol. Either way, yes, raise the original K value to whatever factor you multiply the coefficients by. If the original eqn was 2A + B --> 2C, we can multiply it by 4 to get 8A + 4B --> 8C, and our K that would describe this specific reaction would be K^4

 

[Josh779]

First, I should perhaps clarify what I said earlier. I should have said that when you manipulate the coefficients by multiplying it by a certain factor, you are not really "changing" the equilibrium constant (only temperature changes actually change the K value). Each K value describes a very specific reaction with very specific coefficients. So when we change these coefficients, we are just calculating the K value for that describes the new coefficients. So while I wouldn't necessary call it a different reaction, we are dealing with a different K value.

That probably didn;t make too much sense, lol. Either way, yes, raise the original K value to whatever factor you multiply the coefficients by. If the original eqn was 2A + B --> 2C, we can multiply it by 4 to get 8A + 4B --> 8C, and our K that would describe this specific reaction would be K^4

Doesn't temperature and a catalyst change the rate constant K?

 

[mh0000]

Doesn't temperature and a catalyst change the rate constant K?

rate constant= yes, both catalyst and temp change alters the rate constant. equilibrium constant= no, only temp changes. Think of the disastrous consequences on biological systems if catalysts changed the equilibrium constants.

 

[Josh779]

rate constant= yes, both catalyst and temp change alters the rate constant. equilibrium constant= no, only temp changes. Think of the disastrous consequences on biological systems if catalysts changed the equilibrium constants.

Oh lol, I didn't see equilibrium...proceed haha

What if only the reactants were cut in half? How would that affect K? Would the numerator [P] get cut in half?

 

[mh0000]

Oh lol, I didn't see equilibrium...proceed haha

🙂

 

[rockclock]

First, I should perhaps clarify what I said earlier. I should have said that when you manipulate the coefficients by multiplying it by a certain factor, you are not really "changing" the equilibrium constant (only temperature changes actually change the K value). Each K value describes a very specific reaction with very specific coefficients. So when we change these coefficients, we are just calculating the K value for that describes the new coefficients. So while I wouldn't necessary call it a different reaction, we are dealing with a different K value.

That probably didn;t make too much sense, lol. Either way, yes, raise the original K value to whatever factor you multiply the coefficients by. If the original eqn was 2A + B --> 2C, we can multiply it by 4 to get 8A + 4B --> 8C, and our K that would describe this specific reaction would be K^4

Just take this for granted...that's all you need to know for the DAT: K_new = K_old^n


DISCLAIMER: this isn't important...just if you happen to be curious...

In reality, if your reaction is described by 2A + B --> 2C and that's what happens when you mix A and B in a beaker, then those are the coefficients you should be using for the K value. There's no "special" way of mixing A and B together that would allow you to get them to react as 8A + 4B --> 8C and behave according to a different K value.

Technically this rule only applies to rate laws (small k) for each reaction step (unimolecular, bimolecular, termolecular, etc.) in a mechanism because those relate to the actual physics of atomic collisions. Because equilibrium constants (big K) are a ratio of rate laws (forward and reverse reactions have same rate at equilibrium), this effect carries over into them, as well. Point being, to actually observe K = K^n you would have to be able to alter the molecular mechanism by which A and B react, which obviously you can't. The use of this property is actually the other way around: you can tell things about the reaction mechanism based on how the reaction rates change under different conditions.

Obviously, the nuts and bolts of all that is graduate level chemistry too complicated to test in a general chemistry class or the DAT, so they just make up problems like these so that you can show you understand K = K^n.

http://en.wikipedia.org/wiki/Law_of_mass_action
http://en.wikipedia.org/wiki/Molecularity

 

[Josh779]

Thanks rockclock. Just so I have this straight. To use Kold=knew...If the reaction is reversed then just inverse K and then raise it to the power of the changed coefficients (if they are changing)...correct?

 

[rockclock]

Thanks rockclock. Just so I have this straight. To use Kold=knew...If the reaction is reversed then just inverse K and then raise it to the power of the changed coefficients (if they are changing)...correct?

Yes, sir...

K_new = K_old^n

multiplying all the coefficients by 2: n=2
halving all the coefficients: n=1/2
switching products and reactants: n=-1 ...that -1 power is the same as 1/K
switching products and reactants AND halving all the coefficients: n=-1/2
...
etc.

think of it as if this were a basic math equation instead of a chemical equation
A + 2B = C ..."reactants" A and B, "product" C
multiply everything by -1 (i.e. n=-1)
-A - 2B = -C
C = A + 2B ...switch everything around and now C is the "reacant", A and B "products"

 

[SN1]

Yes, sir...

K_new = K_old^n

multiplying all the coefficients by 2: n=2
halving all the coefficients: n=1/2
switching products and reactants: n=-1 ...that -1 power is the same as 1/K
switching products and reactants AND halving all the coefficients: n=-1/2
...
etc.

think of it as if this were a basic math equation instead of a chemical equation
A + 2B = C ..."reactants" A and B, "product" C
multiply everything by -1 (i.e. n=-1)
-A - 2B = -C
C = A + 2B ...switch everything around and now C is the "reacant", A and B "products"

Just want to make sure. In this problem the reaction was reversed and its coefficient cut in half so

K old = 2
K new = 2^-1/2 which simplifies to (1) / (2^1/2) , is this correct?

Answer A is (1/2.0)^1/2 but I thought only the 2 in the denominator is being raised to 1/2 power NOT the the whole thing.

sorry if this is confusing

 

[rockclock]

Just want to make sure. In this problem the reaction was reversed and its coefficient cut in half so

K old = 2
K new = 2^-1/2 which simplifies to (1) / (2^1/2) , is this correct?

Answer A is (1/2.0)^1/2 but I thought only the 2 in the denominator is being raised to 1/2 power NOT the the whole thing.

sorry if this is confusing

1 raised to any power is 1, and the numerator after taking the reciprocal of K_old is always going to be 1...yes, n = -1/2 and the math works out fine.

come on, i'd expect a 4.0 to be able to handle exponents a little bit better than that haha 😛

 

[SN1]

1 raised to any power is 1, and the numerator after taking the reciprocal of K_old is always going to be 1...yes, n = -1/2 and the math works out fine.

come on, i'd expect a 4.0 to be able to handle exponents a little bit better than that haha 😛

Lol I know 1 raised to any first power is one but the parenthesis in the answer isn't used correctly. That's all lol

 

[rockclock]

Lol I know 1 raised to any first power is one but the parenthesis in the answer isn't used correctly. That's all lol

yes it is.

(1/2.0)^(1/2) = 1/[2.0^(1/2)] = 2.0^(-1/2)

all are the same answer.

the first (answer A) is how you'd get it if you switched reactants and products first and then multiplied everything by 1/2.

the second is how you'd get it if you multiplied everything by 1/2 first and then switched products and reactants.

the third is how you'd get it if you plugged in n = -1/2.

 

[SN1]

yes it is.

(1/2.0)^(1/2) = 1/[2.0^(1/2)] = 2.0^(-1/2)

all are the same answer.

the first (answer A) is how you'd get it if you switched reactants and products first and then multiplied everything by 1/2.

the second is how you'd get it if you multiplied everything by 1/2 first and then switched products and reactants.

the third is how you'd get it if you plugged in n = -1/2.

After writing them on paper you are correct! Sorry for the confusion.

 

[Stacker]

How mnay isomeric mono-substitution products are possible for the bromination of the aromatic hydrocarbon naphthalene?

A. 1
B. 2
C. 3
D. 4
E. 5

That's naphthalene. There are only 2 ways you can mono-brominate that, without getting identical structures.

But those two spots are chiral centers so you have 2^n = 4ways of mono-brominating it. So 4 different products not 2, right?

 

[NDPitch]

But those two spots are chiral centers so you have 2^n = 4ways of mono-brominating it. So 4 different products not 2, right?

Those are not chiral centers!

At a high level, in order for something to be chiral, it has to be attached to 4 different things. Those carbons are not attached to 4 different things. They are only attached to 2 equivalent carbons and a hydrogen.